Question

In a test of braking performance, a tire manufacturer measured the stopping distance for one of its tire models. On a test track, a car made repeated stops from 60 miles per hour. The test was run on both dry and wet pavement, with results as shown below.

Wet   Dry
210   154
191   153
214   143
200   137
202   130
196   146
198   132
179   144
194   142
217   127

a) Construct a 95% confidence interval for the mean dry pavement stopping distance.

The confidence interval is?

(Round to one decimal place as needed.)

b) Construct a 95% confidence interval for the mean increase in stopping distance on wet pavement.

The confidence interval is?

(Round to one decimal place as needed.)

Answer 1

a)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   2.262   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   9.2232/√10=   2.9166          
margin of error , E=t*SE =   2.2622   *   2.9166   =   6.598
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    140.80   -   6.5979   =   134.2021
Interval Upper Limit = x̅ + E =    140.80   -   6.5979   =   147.3979
95%   confidence interval is (   134.2 < µ <   147.4 )

b)

Sample #1   ---->   1
mean of sample 1,    x̅1=   200.100
standard deviation of sample 1,   s1 =    11.387
size of sample 1,    n1=   10
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   140.800
standard deviation of sample 2,   s2 =    9.223
size of sample 2,    n2=   10

Degree of freedom, DF=   n1+n2-2 =    18              
t-critical value =    t α/2 =    2.1009   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    10.3615              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    4.6338              
margin of error, E = t*SE =    2.1009   *   4.63   =   9.74  
                      
difference of means =    x̅1-x̅2 =    200.1000   -   140.800   =   59.30

confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    59.30 -   9.7353   =   49.6
Interval Upper Limit=   (x̅1-x̅2) + E =    59.30 +   9.7353   =   69.0
                      

please revert for doubts and

please UPVOTE the solution